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feat: add solutions to lc problem: No.1262
No.1262.Greatest Sum Divisible by Three
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‎solution/1200-1299/1262.Greatest Sum Divisible by Three/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：动态规划**
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我们定义 $f[j]$ 表示前 $i-1$ 个数中，余数为 $j$ 的最大和。那么对于当前的数 $x$，我们可以将其加入到前面的数中，得到的和为 $f[j] + x$，其中 $j$ 为 $(f[j] + x) \bmod 3$。
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最后，我们返回 $f[0]$ 即可。
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时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组 `nums` 的长度。
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<!-- tabs:start -->
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### **Python3**
@@ -55,13 +63,13 @@
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```python
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class Solution:
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def maxSumDivThree(self, nums: List[int]) -> int:
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dp = [0] * 3
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for v in nums:
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a, b, c = dp[0] + v, dp[1] + v, dp[2] + v
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dp[a % 3] = max(dp[a % 3], a)
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dp[b % 3] = max(dp[b % 3], b)
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dp[c % 3] = max(dp[c % 3], c)
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return dp[0]
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f = [0] * 3
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for x in nums:
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a, b, c = f[0] + x, f[1] + x, f[2] + x
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f[a % 3] = max(f[a % 3], a)
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f[b % 3] = max(f[b % 3], b)
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f[c % 3] = max(f[c % 3], c)
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return f[0]
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```
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### **Java**
@@ -71,18 +79,58 @@ class Solution:
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```java
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class Solution {
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public int maxSumDivThree(int[] nums) {
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int[] dp = new int[3];
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for (int v : nums) {
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int a = dp[0] + v, b = dp[1] + v, c = dp[2] + v;
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dp[a % 3] = Math.max(dp[a % 3], a);
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dp[b % 3] = Math.max(dp[b % 3], b);
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dp[c % 3] = Math.max(dp[c % 3], c);
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int[] f = new int[3];
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for (int x : nums) {
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int a = f[0] + x, b = f[1] + x, c = f[2] + x;
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f[a % 3] = Math.max(f[a % 3], a);
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f[b % 3] = Math.max(f[b % 3], b);
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f[c % 3] = Math.max(f[c % 3], c);
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}
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return dp[0];
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return f[0];
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int maxSumDivThree(vector<int>& nums) {
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int f[3]{};
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for (int x : nums) {
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int a = f[0] + x, b = f[1] + x, c = f[2] + x;
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f[a % 3] = max(f[a % 3], a);
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f[b % 3] = max(f[b % 3], b);
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f[c % 3] = max(f[c % 3], c);
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}
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return f[0];
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}
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};
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```
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### **Go**
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```go
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func maxSumDivThree(nums []int) int {
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f := [3]int{}
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for _, x := range nums {
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a, b, c := f[0]+x, f[1]+x, f[2]+x
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f[a%3] = max(f[a%3], a)
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f[b%3] = max(f[b%3], b)
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f[c%3] = max(f[c%3], c)
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}
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return f[0]
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **...**
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```

‎solution/1200-1299/1262.Greatest Sum Divisible by Three/README_EN.md

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```python
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class Solution:
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def maxSumDivThree(self, nums: List[int]) -> int:
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dp = [0] * 3
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for v in nums:
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a, b, c = dp[0] + v, dp[1] + v, dp[2] + v
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dp[a % 3] = max(dp[a % 3], a)
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dp[b % 3] = max(dp[b % 3], b)
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dp[c % 3] = max(dp[c % 3], c)
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return dp[0]
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f = [0] * 3
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for x in nums:
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a, b, c = f[0] + x, f[1] + x, f[2] + x
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f[a % 3] = max(f[a % 3], a)
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f[b % 3] = max(f[b % 3], b)
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f[c % 3] = max(f[c % 3], c)
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return f[0]
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```
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### **Java**
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```java
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class Solution {
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public int maxSumDivThree(int[] nums) {
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int[] dp = new int[3];
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for (int v : nums) {
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int a = dp[0] + v, b = dp[1] + v, c = dp[2] + v;
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dp[a % 3] = Math.max(dp[a % 3], a);
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dp[b % 3] = Math.max(dp[b % 3], b);
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dp[c % 3] = Math.max(dp[c % 3], c);
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int[] f = new int[3];
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for (int x : nums) {
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int a = f[0] + x, b = f[1] + x, c = f[2] + x;
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f[a % 3] = Math.max(f[a % 3], a);
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f[b % 3] = Math.max(f[b % 3], b);
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f[c % 3] = Math.max(f[c % 3], c);
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}
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return dp[0];
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return f[0];
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int maxSumDivThree(vector<int>& nums) {
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int f[3]{};
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for (int x : nums) {
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int a = f[0] + x, b = f[1] + x, c = f[2] + x;
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f[a % 3] = max(f[a % 3], a);
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f[b % 3] = max(f[b % 3], b);
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f[c % 3] = max(f[c % 3], c);
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}
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return f[0];
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}
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};
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```
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### **Go**
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```go
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func maxSumDivThree(nums []int) int {
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f := [3]int{}
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for _, x := range nums {
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a, b, c := f[0]+x, f[1]+x, f[2]+x
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f[a%3] = max(f[a%3], a)
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f[b%3] = max(f[b%3], b)
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f[c%3] = max(f[c%3], c)
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}
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return f[0]
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **...**
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```

‎solution/1200-1299/1262.Greatest Sum Divisible by Three/Solution.cpp

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class Solution {
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public:
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int maxSumDivThree(vector<int>& nums) {
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int f[3]{};
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for (int x : nums) {
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int a = f[0] + x, b = f[1] + x, c = f[2] + x;
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f[a % 3] = max(f[a % 3], a);
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f[b % 3] = max(f[b % 3], b);
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f[c % 3] = max(f[c % 3], c);
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}
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return f[0];
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}
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};

‎solution/1200-1299/1262.Greatest Sum Divisible by Three/Solution.go

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func maxSumDivThree(nums []int) int {
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f := [3]int{}
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for _, x := range nums {
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a, b, c := f[0]+x, f[1]+x, f[2]+x
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f[a%3] = max(f[a%3], a)
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f[b%3] = max(f[b%3], b)
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f[c%3] = max(f[c%3], c)
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}
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return f[0]
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}

‎solution/1200-1299/1262.Greatest Sum Divisible by Three/Solution.java

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class Solution {
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public int maxSumDivThree(int[] nums) {
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int[] dp = new int[3];
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for (int v : nums) {
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int a = dp[0] + v, b = dp[1] + v, c = dp[2] + v;
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dp[a % 3] = Math.max(dp[a % 3], a);
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dp[b % 3] = Math.max(dp[b % 3], b);
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dp[c % 3] = Math.max(dp[c % 3], c);
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int[] f = new int[3];
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for (int x : nums) {
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int a = f[0] + x, b = f[1] + x, c = f[2] + x;
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f[a % 3] = Math.max(f[a % 3], a);
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f[b % 3] = Math.max(f[b % 3], b);
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f[c % 3] = Math.max(f[c % 3], c);
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}
10-
return dp[0];
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return f[0];
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}
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}
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}

‎solution/1200-1299/1262.Greatest Sum Divisible by Three/Solution.py

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class Solution:
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def maxSumDivThree(self, nums: List[int]) -> int:
3-
dp = [0] * 3
4-
for v in nums:
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a, b, c = dp[0] + v, dp[1] + v, dp[2] + v
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dp[a % 3] = max(dp[a % 3], a)
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dp[b % 3] = max(dp[b % 3], b)
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dp[c % 3] = max(dp[c % 3], c)
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return dp[0]
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f = [0] * 3
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for x in nums:
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a, b, c = f[0] + x, f[1] + x, f[2] + x
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f[a % 3] = max(f[a % 3], a)
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f[b % 3] = max(f[b % 3], b)
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f[c % 3] = max(f[c % 3], c)
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return f[0]

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